indefiniteintegral
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- Mar 8, 2020
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So I'm currently, taking Calculus 2 and I'm on Series right now and in 11.2 it's the first chapter on series, and covers the Divergence test,
In the first set of problems it wants you to find if the series in convergent, and if so provide the sum
So here is the problem I was solving
[math]\displaystyle\sum_{i=2}^n\frac{2}{n^2-1}[/math]
And I see that if I take the limit of the series at infinity it goes to zero, which means its inconclusive as to if it converges or not.
So another test must be used.
So me being bored this weekend learned the integral test before any other test, and when I saw this problem , I thought just my luck!
I will use partial fractions, solve it as an improper integral, get the limit and I'm onto the next problem hooray for me easy peasy
So I set my f(x) as follow:
[math]\int^∞_2 \frac{2}{x^2-1}[/math]
[imath]=\int^∞_2 \frac{2}{x^2-1}= \int^∞_2 \frac{A}{x-1} +\frac{B}{x+1}[/imath]
[imath]=\int^∞_2 \frac{1}{x-1} -\frac{1}{x+1}[/imath]
[imath]= ln| \frac{x-1}{x+1}|[/imath]
So now I can take this and evaluate the limit from 2 to t as t goes to infinity
Which then leaves me with
[imath]= 0 - ln \frac{1}{3}[/imath]
[imath]= ln|3|[/imath]
So there was my final answer. But I looked it up in the textbook, and they "Used the definition of a convergent series and used partial fractions to get the partial sums"
They used [imath]\displaystyle\sum_{i=2}^n\frac{1}{i-1} - \frac{1}{i+1}[/imath]
From there they used a telescoping series to get a final answer which is not the answer I got.
So I'm sorry for typing so much, and I apologize if my question is stupid but my question is, if the sum of the series has to be calculated the way specifically done by the book, then what on earth did I calculate by using the integral test? Did I not calculate the sum of the function with the integral test? Is the sum of the function not the same as the sum of the series? I feel like I am fundamentally missing something here.
Thanks for any help!
In the first set of problems it wants you to find if the series in convergent, and if so provide the sum
So here is the problem I was solving
[math]\displaystyle\sum_{i=2}^n\frac{2}{n^2-1}[/math]
And I see that if I take the limit of the series at infinity it goes to zero, which means its inconclusive as to if it converges or not.
So another test must be used.
So me being bored this weekend learned the integral test before any other test, and when I saw this problem , I thought just my luck!
I will use partial fractions, solve it as an improper integral, get the limit and I'm onto the next problem hooray for me easy peasy
So I set my f(x) as follow:
[math]\int^∞_2 \frac{2}{x^2-1}[/math]
[imath]=\int^∞_2 \frac{2}{x^2-1}= \int^∞_2 \frac{A}{x-1} +\frac{B}{x+1}[/imath]
[imath]=\int^∞_2 \frac{1}{x-1} -\frac{1}{x+1}[/imath]
[imath]= ln| \frac{x-1}{x+1}|[/imath]
So now I can take this and evaluate the limit from 2 to t as t goes to infinity
Which then leaves me with
[imath]= 0 - ln \frac{1}{3}[/imath]
[imath]= ln|3|[/imath]
So there was my final answer. But I looked it up in the textbook, and they "Used the definition of a convergent series and used partial fractions to get the partial sums"
They used [imath]\displaystyle\sum_{i=2}^n\frac{1}{i-1} - \frac{1}{i+1}[/imath]
From there they used a telescoping series to get a final answer which is not the answer I got.
So I'm sorry for typing so much, and I apologize if my question is stupid but my question is, if the sum of the series has to be calculated the way specifically done by the book, then what on earth did I calculate by using the integral test? Did I not calculate the sum of the function with the integral test? Is the sum of the function not the same as the sum of the series? I feel like I am fundamentally missing something here.
Thanks for any help!